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2. Manipulator Kinematics 2. MANIPULATOR KINEMATICS Position vectors and their transformations Direct and inverse kinematics of manipulators Transformation of velocity and torque vectors Classification of kinematical chains of manipulator Cartesian, polar cylindrical and spherical and angular coordinates of manipulators Multilink manipulators and manipulators with flexible links Manipulators with parallel kinematics Kinematics of mobile robots 2.1. Position vectors and their transformation Manipulator kinematics is the field of science that investigates the motion of manipulator links without regard to the forces that cause it. In that case the motion is determined with trajectory, i.e. positions, velocity, acceleration, jerk and other higher derivative components. r The location of a point A in Cartesian coordinates {A} is determined by vector AP (Fig 2.1), the individual element of which are projections to the three orthogonal axes of coordinates x, y, z, respectively. The projections px, py and pz can be considered as three orthogonal vectors. ⎡ px ⎤ r AP = ⎢ p ⎥. (2.1) ⎢ y ⎥ ⎣⎢ pz ⎦⎥ The orientation of a body in Cartesian space {A} is determined by three unit vectors r r r ,, zyx BBB that are attached to the body. These orthogonal unit vectors form the Cartesian coordinate system {B} that can be called body coordinates {B}. Therefore, the orientation of a body in space coordinates can be described by the rotation of body coordinates {B} relative r to the reference coordinates {A}. The unit vector xB can be described in Cartesian space coordinates {A} with three projections to coordinate axes xA, yA, zA. In the same way, other r r unit vectors yB and zB can be described. {A} z pz AP OA y py px x Figure 2.1. Position vector of a point in Cartesian coordinates 32 Generally X X X ⎡ xB ⎤ ⎡ yB ⎤ ⎡ zB⎤ r ⎢ ⎥ r ⎢ ⎥ r ⎢ ⎥ A Y A Y A Y . (2.2) X B = ⎢ xB ⎥ YB = ⎢ yB ⎥ ZB = ⎢ zB ⎥ ⎢ Z ⎥ ⎢ Z ⎥ ⎢ Z ⎥ ⎣ xB ⎦ ⎣ yB ⎦ ⎣ zB ⎦ A r A r A r A Three vectors: X B , YB and ZB will form the rotation matrix B R, which describes the orientation of coordinates {B} relative to coordinates {A}. X X X ⎡ xB yB zB ⎤ A ⎢ Y Y Y ⎥ A rA rA r B R = ⎢ xB yB zB ⎥ = [XYZB,,.B B ] (2.3) ⎢ Z Z Z ⎥ ⎣ xB yB zB ⎦ The position of coordinates {B} relative to coordinates {A} (Fig. 2.2) is defined by the A r position vector PBO of the origin of coordinates {B} in coordinates {A} and rotation matrix A B R. AA r {}B,.= {B RPBO } (2.4) r The position vector B P in translated coordinates {B} relative to coordinates {A} (Fig. 2.3) is defined by the sum of two vectors ABAr r r PPP= + BO . (2.5) r The position vector B P in rotated coordinates {B} relative to coordinates {A} is defined A A r by the rotation matrix B R . In this case the position vector P is determined as the scalar A B r product of the rotation matrix B R and the position vector P (2.6). {A} zA {B} xB OB yB A zB PBO OA yA xA Figure 2.2 Determination of manipulator gripper or tool orientation 33 {Β} zB {Α} zA BP A A P PBO OB yB xB OA yA x A Figure 2.3 Position vectors in translated coordinates {A} and {B} A rAB r PRP=B ⋅ . (2.6) A The rotation matrix B R describes the rotation of coordinates {B} relative to coordinates {A}. We can use the rotation matrix to describe inversely coordinates {A} relative to coordinates B {B}. This transformation will be described by the rotation matrix A R . It is known from the course of linear algebra that the inverse of matrix with orthonormal columns is equal to its transpose. The transpose of the matrix is defined by the change of elements of rows and columns. B A −1 AT A RRR=B = B . (2.7) The program MathCAD Symbolics helps us to calculate and we can easily demonstrate that A −1 AT the inverse of the rotation matrix is equal to its transpose. B RR= B , because the nominator of elements in rotation matrix R−1 is equal to one. ⎛ cos()α −sin()α 0 ⎞ R := ⎜ ⎟ ⎜ sin()α cos()α 0 ⎟ ⎝ 0 0 1 ⎠ ⎛ cos()α sin()α 0 ⎞ T ⎜ ⎟ R → ⎜ −sin()α cos()α 0 ⎟ ⎜ ⎟ ⎝ 0 0 1 ⎠ ⎡ cos()α sin()α ⎤ 0 ⎢ ()cos()α 2 + sin()α 2 ()cos()α 2 + sin()α 2 ⎥ ⎢ ⎥ − 1 R → ⎢ −sin()α cos()α ⎥ 0 ⎢ ()cos()α 2 + sin()α 2 ()cos()α 2 + sin()α 2 ⎥ ⎢ ⎥ ⎣ 0 0 1 ⎦ 34 The matrix with orthogonal elements the norm of which is equal to 1 (orthogonal unit vectors) is called orthonormal. The transposed matrix is the matrix with transposed rows and columns. The inverse matrix is calculated by the formula 1 R−1 = (R ), (2.8) det R ji where Rji is an adjunct matrix. The inverse matrix of the orthonormal matrix is equal to the transposed matrix. Because the rotation matrix is orthonormal, its inverse matrix can be found as its transpose. The product of the matrix and its inverse gives a unit matrix. The unit matrix has diagonal elements equal to 1, and all other elements equal to 0. The scalar product of matrixes can be calculated by using the formula: n cabik =⋅∑(ij ik ). (2.9) j=1 The formula can be used for the calculation of product 2.6. In this case the matrix aij is the B r A r rotation matrix and bjk is the position vector P . The result cik is the position vector P , where (i = 1...3; j = 1...3; k = 1). r The general transform of the position vector, the vector B P given in translated and rotated coordinates {B} in the relation of base coordinates {A} can be described by the formula A rABA r r PRPP=B ⋅ + BO . (2.10) r The vector AP can also be calculated by the transformation matrix A r AB r PTP=B ⋅ , (2.11) A where BT is the transformation matrix to transform translated and rotated coordinates {B} to the base coordinates {A}. The procedure is called as the Homogeneous Transform. The transformation matrix to transform coordinates {B} includes 3x3 rotation matrix and 3x1 position vector of origin coordinates {B} in the relation of coordinates {A}. The formula 2.11 can be shown as follows: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ AP⎥ ⎢ ARPPA ⎥ ⎢ B ⎥ ⎢ ⎥ = ⎢ B BO ⎥ ⋅ ⎢ ⎥. (2.12) ⎢ _ ⎥ ⎢____ ⎥ ⎢ _ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 1 ⎦ ⎣0 0 0 1 ⎦ ⎣ 1 ⎦ Addition of the last row [0 0 0 1] is a mathematic manipulation to get 4x4 square matrix. 35 The use of transformation matrixes (transformation operators) helps us to describe different spatial vectors in different coordinates. For example, we can describe the motion of multi link manipulators and to solve the direct and inverse kinematics tasks of manipulators. The general transformation can be replaced by separate translation and rotation transforms and the corresponding transform operators. ⎡1 0 0 qX ⎤ ⎢0 1 0 q ⎥ TRANS(,) Q Q = ⎢ Y ⎥ (2.13) ⎢0 0 1 qZ ⎥ ⎢ ⎥ ⎣0 0 0 1 ⎦ where Q is the vector and Q its module. ⎡cosα− sin α 0 0⎤ ⎢sinα cos α 0 0⎥ ROT(,) Z α = ⎢ ⎥, (2.14) ⎢ 0 0 1 0⎥ ⎢ ⎥ ⎣ 0 0 0 1⎦ where Z defines the rotation around z-axis and α the rotation angle. The separate translation and rotation operators can be combined in one transformation operator (transformation matrix). Several sequential transforms can be combined in one. The transform equations: B rBC r A r AB r PTP=C ⋅ and PTP=B ⋅ can be combined in the following equation: A r A BC r PTTP=B ⋅C ⋅ . (2.15) Transformation matrix for the combined transform ⎡ ⎤ r r ⎢ ARRRPP⋅ B AB⋅ + A ⎥ AT = ⎢ B C B CO BO ⎥. (2.16) C ⎢____ ⎥ ⎢ ⎥ ⎣0 0 0 1 ⎦ B Inverse transform matrix AT can be presented as follows: ⎡ ⎤ r ⎢ ATR −ATARP ⋅ ⎥ BT = ⎢ B B OB ⎥ (2.17) A ⎢____ ⎥ ⎢ ⎥ ⎣0 0 0 1 ⎦ B A −1 or ATT= B . 36 B The transformation matrix AT is not orthonormal (consisting only of orthogonal unit vectors) and therefore its inverse matrix is not equal to the transposed matrix. By solving transform equations it is possible to find solutions for any unknown transform if sufficient transform conditions exist. For example, if we know the conditions E E A E E B C DTTT=A ⋅ D and DTTTT=B ⋅C ⋅ D E A E B C then ATTTTT=D =B ⋅C ⋅ D . Consequently, B E −1 A C −1 CTTTT=B ⋅D ⋅ D . (2.18) Euler angles. The rotation matrix is not the only way to define the orientation of a body. We can define the orientation of a body also by three independent angles. The nine elements of the rotation matrix are mutually dependent and there are six conditions, including three conditions for the longitude of the unit vector Xˆ = ;1 Yˆ = ;1 Zˆ = ;1 (2.19) and three conditions for their orthogonal location exist. XYˆ⋅ ˆ = ;0 XZˆ⋅ ˆ = ;0 YZˆ⋅ ˆ = 0 (2.20) The following calculation using the MathCAD program illustrates the issue of scalar or dot product and cross product of orthogonal vectors. The dot product of two orthogonal or parallel vectors with magnitude 5 gives ⎜⎛ 0 ⎞⎟ ⎜⎛ 5 ⎞⎟ ⎜⎛ 5 ⎟⎞ ⎜⎛ 5 ⎞⎟ ⎜ 5 ⎟⋅⎜ 0 ⎟ = 0 ⎜ 0 ⎟⋅⎜ 0 ⎟ = 25 ⎜ 0 ⎟ ⎜ 0 ⎟ ⎜ 0 ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ r The cross product of the same vectors ar × b = nr ⋅ a ⋅ b ⋅ sinα gives ⎜⎛ 5 ⎟⎞ ⎜⎛ 0 ⎟⎞ ⎜⎛ 0 ⎟⎞ ⎜⎛ 5 ⎟⎞ ⎜⎛ 5 ⎟⎞ ⎜⎛ 0 ⎟⎞ ⎜ 0 ⎟ × ⎜ 5 ⎟ = ⎜ 0 ⎟ ⎜ 0 ⎟ × ⎜ 0 ⎟ = ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 25 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ , r where the resultant vector nr is orthogonal to the unit vector ar and b .
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